Some chosen topics on Advanced Algebra, simplified

1. FACTOR THEOREM
Long Division,
Synthetic Division,
Remainder Theorem
&

2. INTRODUCTION
In Advance Algebra, one of the most
interesting topics are dividing
Polynomials. Dividing is one of the most
confusing operations among students.
This presentation aims to teach us the
2 ways to divide polynomials and with 2
bonus topics: The Remainder Theorem
and The Factor Theorem. This
presentation is about to help most of us in
dividing polynomials from a very
complicated and confusing operation to an
easy task.

3. TABLE OF CONTENTS
Title
Page………………………………….Slide
1
Introduction………………………………Sli
de 2
Table of Contents………………………Slide
3
2 Ways to Divide Polynomials…….Slides 4-
8
Long Division Method……..........Slides 5-
6

5. FACTOR THEOREM
Long Division Method
Synthetic Division
Remainder Theorem
&
 Is the usual way of dividing the polynomials that was
introduced in Elementary Algebra
STEPS:
1) Divide the first term of both the divisor and dividend.
2) Then, multiply the answer to the divisor. Put the product
below the Dividend. Just be sure to align it properly.
3) Subtract and bring down the next digit or term.
4) Then, divide again the same way as before.
5) Repeat steps 2-4 until you get 0 or a remainder.
q(x) – quotient
r(x) - remainder

6. GIVEN: Use Long Division Method to divide 7x+3 to 14x2+20x+49
SOLUTION:
2x + 2
7x+3 14x2+20x+49 q(x) = 2x+2
-14x2+6x
14x+49 r(x) = 43
-14x+6
43
1) We divide 7x from 4x2 and we got 2x.
2) We multiply 2x to 7x+3.
3) We subtract, so we change signs (indicated by red
marks).
4) The difference is 14x and we bring down 49.
5) We divide 7x from 14x and we got 2.
6) We then multiply 2 to 7x+3 and subtract it the same as in
step 3.
7) Lastly, we got the remainder 43 and quotient 2x+2.

7.  Is another way of dividing polynomials by which we only use the
constants among the given functions
FACTOR THEOREM
Long Division
Synthetic Division
Remainder Theorem
&
STEPS:
1) First, get the constants in the function as shown
below:
from x+4 8x2+6x+2 to 4 8 6 2
2) Bring down the first digit.
3) Multiply it to the divisor. Put the product below the
next digit.
4) Subtract.
5) Repeat steps 2-4 until done.
6) Add xn-1 on the digits. Remember that the last digit
would be the remainder. For instance, the answer for
12x2-15x+3 / x-2 is 12 9 21. Then, the final answer
should be: q(x) = 12x+9 & r(x) = 21.

8. GIVEN: Use Synthetic division to divide x+2 from 4x4+2x3+x+5
SOLUTION:
2 4 2 0 1 5 q(x) = 4x3-6x2+12x-23
(-)8 -12 (-)24 -46 r(x) = 51
4 -6 12 -23 51
1) Since we don’t have a term with x2, we replaced it by 0 to
complete the function.
2) We bring down 4.
3) We multiply it to 2 and we got 8.
4) We subtract, so we change signs.
5) We did the same to the next digits.
6) The final answer is 4x3-6x2+12x+23 and the remainder is 51.
To prove that we are correct, we would use the Long Division method to solve the
same problem….. .. .. ..
4x3-6x2+12x-23
x+2 4x4+2x3+x+5 q(x) = 4x3-6x2+12x-23
4x4+8x3
-6x3+x r(x) = 51
-6x3+12x2
12x2+x *We used 2 different
12x2+24x methods but we
-23x+5 arrived w/ the same
-23x-46 answer.
51

9. FACTOR THEOREM
Long Division
Synthetic Division
Remainder Theorem
&
 It is simply a way in finding the remainder of a polynomial
equation w/o finding its quotient
STEPS:
1) Get the inverse of the constant of the divisor. This
would be the value of x.
2) Substitute it to the dividend.
3) Solve. Perform the indicated operations.
4) The answer would be the remainder.

10. GIVEN: Consider the given in Synthetic division:
SOLUTION:
X = -2
4(-2)4+2(-2)3-2+5
4(16)+29(-8)+3
64-16+3
67-16
= 51
*Looking back to slide 8, the
remainder were the same.
1) First, we get the inverse of constant of x+2 which is -2. It
would be the value of x.
2) We substitute -2 to all x of the dividend.
3) We first solve the exponents and multiply.
4) Add the digits with the same sign.
5) Lastly, we add them and we got 51, the same remainder as
we use Long and Synthetic Division Method.

11. FACTOR THEOREM
Long Division,
Synthetic Division,
Remainder Theorem
&
 It is just a way of checking if the divisor is a factor of the
dividend
STEPS:
1) Multiply the divisor to the quotient.
2) Then, add the remainder (if any).
3) The answer should be the dividend. If not, then the
divisor is not a factor of the dividend.

12. GIVEN: Consider the previous given (in Synthetic Division)
SOLUTION:
q(x) = 4x3-6x2+12x-23 divisor = x+2
r(x) = 51 dividend = 4x4+2x3+x+5
4x3-6x2+12x-23  q(x)
x x+2  divisor
4x4-6x3+12x2-23x
8x3-12x2+24x-46
4x4+2x3+x-46
+ 51  r(x)
4x4+2x3+x+5  dividend
1) We first multiply q(x) and the divisor.
Then, we add r(x).
2) The answer is the dividend.
Therefore, x+2 is a factor of 4x4+2x3+x+5.

13. GENERALIZATION
There are 2 ways to divide polynomials: the Long Division Method
and The Synthetic Division.
The Long Division Method is the way of dividing polynomials that
was taught in our 1st and 2nd year Algebra.
The Synthetic Division is much easier and faster way in dividing
polynomials.
Remainder Theorem focuses on finding only the remainder even if
not divided.
Factor Theorem is just like checking operation in division of
numbers.

14. Here is a table that contains different ways of dividing polynomials
that includes the Remainder Theorem and The Factor Theorem.
GIVEN: 4x4+2x3+x+5 / x+2
LONG DIVISION
METHOD
SYNTHETIC
DIVISION
REMAINDER
THEOREM
FACTOR
THEOREM
q(x) =
4x3-6x2+12x-
23
r(x) = 51
X = -2
4(-2)4+2(-2)3-
2+5
4(16)+29(-8)+3
64-16+3
67-16
= 51
4x3-6x2+12x-23
x x+2
4x4-6x3+12x2-23x
8x3-12x2+24x-46
4x4+2x3+x-46
+ 51
4x4+2x3+x+5
Therefore, x+2 is
a factor of
4x4+2x3+x+5.
2 4 2 0 1 5
(-)8 -12 (-)24 -46
4 -6 12 -23 51
4x3-6x2+12x-23
x+2 4x4+2x3+x+5
4x4+8x3
-6x3+x
-6x3+12x2
12x2+x
12x2+24x
-23x+5
-23x-46
51
q(x) = 4x3-6x2+12x-23
r(x) = 51

15. JOHN ROME R. ARANAS
Creator
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New Century Mathematics
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Ms. Charmaigne Marie Mahamis
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